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Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts a force of 69.0 N, the second a force of 91.0 N, friction is 15.0 N, and the mass of the third child plus wagon is 23.0 kg.
What is the system of interest if the acceleration of the child in the wagon is to be calculated?

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MechEngineer

Answer:

0.304 m/s2

Explanation:

If the first child is pushing with a force of 69N to the right and the 2nd child is pushing with a force of 91N to the left. Then the net pushing force is 91 - 69 = 22 N to the left. Subtracted by 15N friction force then the system of interest is subjected to F = 7 N net force tot he left.

We can use Newton's 2nd law to calculate the net acceleration of the system

[tex]a = F/m = 7 / 23 = 0.304 m/s^2[/tex]

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